## Full text: Pergaeus, Apollonius: The two books of Apollonius Pergaeus, concerning tangencies, as they have been restored by Franciscus Vieta and Marinus Ghetaldus

and P will be greater than one to R and half AE, and of courſe, AQ
(a fourth proportional to S, R and P) greater than a third proportional
to P and half AE; in which caſe the rectangle AQ, P will be greater
than the ſquare on half AE, and ſo AD (a mean proportional between
AQ and P) greater than half AE; but when this happens, it is plain that
DH can neither cut nor touch the circle on AE, and therefore, the
problem becomes impoſſible.

## 73.PROBLEM III. (Fig. 4. and 5.)

In any indefinite ſtraight line let there be aſſigned the points A and E; it is required to cut it in another point O, ſo that the ſquare on the ſegment
AO may be to the rectangle contained by the ſegment EO and a given line
P, in the ratio of two given ſtraight lines R and S.

Analysis . Suppoſe the thing done, and that O is the point ſought: then will the ſquare on AO be to the rectangle EO, P as R to S. Make
AQ to P as R is to S; then will the ſquare on AO be to the rectangle EO,
P as AQ is to P; or ( Eu . V. 15.) the ſquare on AO is to the rectangle EO,
P as the rectangle AQ, AO is to the rectangle P, AO; wherefore AO is to
EO as AQ to AO; conſequently by compoſition, or diviſion, AO is to AE
as AQ is to OQ, and ſo ( Eu . VI. 16.) the rectangle AO, OQ is equal to
the rectangle AE, AQ; and hence, as the ſum or difference of AO and OQ
is alſo given, theſe lines themſelves are given by the 85th or 86th of the
Data.

Synthesis . Take AQ a fourth proportional to S, P and R, and
deſcribe thereon a circle; erect at A, the indefinite perpendicular AK, and
take therein AD, a mean proportional between AE and AQ; from D,
draw DH, parallel to AE, if O be required beyond E; but through F the
center of the circle on AQ, if it be ſought beyond A, or between A and
E, cutting the ſaid circle in H: Laſtly, from H draw HO perpendicular
to DH, which will cut the indefinite line in O, the point required.

For it is plain from the Conſtruction, that AD and HO are equal; and
( Eu . VI. 17) the rectangle AE, AQ is equal to the ſquare on AD, and
therefore equal to the ſquare on HO; but the ſquare on HO is equal to the
rectangle AO, OQ, ( Eu . III. 35. 36) conſequently the rectangle AO, OQ

### Note to user

Dear user,

In response to current developments in the web technology used by the Goobi viewer, the software no longer supports your browser.